/*
 * Copyright (C) 2008 Google Inc.
 *
 * Licensed under the Apache License, Version 2.0 (the "License");
 * you may not use this file except in compliance with the License.
 * You may obtain a copy of the License at
 *
 * http://www.apache.org/licenses/LICENSE-2.0
 *
 * Unless required by applicable law or agreed to in writing, software
 * distributed under the License is distributed on an "AS IS" BASIS,
 * WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
 * See the License for the specific language governing permissions and
 * limitations under the License.
 */

package com.itranswarp.javapractice.plugin.repackaged.com.google.gson;

import java.lang.reflect.Type;

/**
 * <p>
 * Interface representing a custom deserializer for Json. You should write a
 * custom deserializer, if you are not happy with the default deserialization
 * done by Gson. You will also need to register this deserializer through
 * {@link GsonBuilder#registerTypeAdapter(Type, Object)}.
 * </p>
 *
 * <p>
 * Let us look at example where defining a deserializer will be useful. The
 * {@code Id} class defined below has two fields: {@code clazz} and
 * {@code value}.
 * </p>
 *
 * <pre>
 * public class Id&lt;T&gt; {
 * 	private final Class&lt;T&gt; clazz;
 * 	private final long value;
 * 
 * 	public Id(Class&lt;T&gt; clazz, long value) {
 * 		this.clazz = clazz;
 * 		this.value = value;
 * 	}
 * 
 * 	public long getValue() {
 * 		return value;
 * 	}
 * }
 * </pre>
 *
 * <p>
 * The default deserialization of {@code Id(com.foo.MyObject.class, 20L)} will
 * require the Json string to be
 * <code>{"clazz":com.foo.MyObject,"value":20}</code>. Suppose, you already know
 * the type of the field that the {@code Id} will be deserialized into, and
 * hence just want to deserialize it from a Json string {@code 20}. You can
 * achieve that by writing a custom deserializer:
 * </p>
 *
 * <pre>
 * class IdDeserializer implements JsonDeserializer&lt;Id&gt;() {
 *   public Id deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context)
 *       throws JsonParseException {
 *     return new Id((Class)typeOfT, id.getValue());
 *   }
 * </pre>
 *
 * <p>
 * You will also need to register {@code IdDeserializer} with Gson as follows:
 * </p>
 *
 * <pre>
 * Gson gson = new GsonBuilder().registerTypeAdapter(Id.class, new IdDeserializer()).create();
 * </pre>
 *
 * <p>
 * New applications should prefer {@link TypeAdapter}, whose streaming API is
 * more efficient than this interface's tree API.
 *
 * @author Inderjeet Singh
 * @author Joel Leitch
 *
 * @param <T> type for which the deserializer is being registered. It is
 *        possible that a deserializer may be asked to deserialize a specific
 *        generic type of the T.
 */
public interface JsonDeserializer<T> {

	/**
	 * Gson invokes this call-back method during deserialization when it encounters
	 * a field of the specified type.
	 * <p>
	 * In the implementation of this call-back method, you should consider invoking
	 * {@link JsonDeserializationContext#deserialize(JsonElement, Type)} method to
	 * create objects for any non-trivial field of the returned object. However, you
	 * should never invoke it on the the same type passing {@code json} since that
	 * will cause an infinite loop (Gson will call your call-back method again).
	 *
	 * @param json    The Json data being deserialized
	 * @param typeOfT The type of the Object to deserialize to
	 * @return a deserialized object of the specified type typeOfT which is a
	 *         subclass of {@code T}
	 * @throws JsonParseException if json is not in the expected format of
	 *                            {@code typeofT}
	 */
	public T deserialize(JsonElement json, Type typeOfT, JsonDeserializationContext context) throws JsonParseException;
}
